如何使用“尝试捕获”在颤音中进行错误处理
我有一个API端点,当我插入重复条目时返回错误响应。我需要使用尝试捕获量捕获此错误响应,以便将此响应作为吐司消息传递。这是我尝试过的API响应结构和API消耗代码:
// companyService
Future<Object?> createCompany(Company company) async {
try {
var content = jsonEncode(company.toJson());
accessToken = await sharedPref.read(key);
searchUrl = "$baseUrl/company";
var response = await http.post(
Uri.parse(searchUrl),
body: content,
headers: {
"Content-Type": "application/json",
"Authorization": "Bearer $accessToken",
},
);
if (response.statusCode == 201) {
return company;
} else {
return null;
}
} catch (e) {
print('objectError: $e');
return e;
}
// companyFactory
createCompany(
name,
domain,
address,
cityId,
) async {
final company = Company(
name: name,
domain: domain,
address: address,
cityId: cityId,
);
try {
CompanyService requests = CompanyService();
await requests.createCompany(company);
return company;
} catch (e) {
print('objectError: $e');
return e;
}
}
I have an API endpoint that returns an error response when I insert a duplicate entry. I need to capture this error response using the try catch so I can pass this response as a toast message. Here is the api response structure and the api consumption code I have tried:
// companyService
Future<Object?> createCompany(Company company) async {
try {
var content = jsonEncode(company.toJson());
accessToken = await sharedPref.read(key);
searchUrl = "$baseUrl/company";
var response = await http.post(
Uri.parse(searchUrl),
body: content,
headers: {
"Content-Type": "application/json",
"Authorization": "Bearer $accessToken",
},
);
if (response.statusCode == 201) {
return company;
} else {
return null;
}
} catch (e) {
print('objectError: $e');
return e;
}
// companyFactory
createCompany(
name,
domain,
address,
cityId,
) async {
final company = Company(
name: name,
domain: domain,
address: address,
cityId: cityId,
);
try {
CompanyService requests = CompanyService();
await requests.createCompany(company);
return company;
} catch (e) {
print('objectError: $e');
return e;
}
}
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